Variance of n natural number is given by,

Variance$=\frac{\sum n^2}{n}-{\left(\frac{\sum n}{n}\right)}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6n}-{\left(\frac{n\left(n+1\right)}{2n}\right)}^2$

                  $=\left(n+1\right)\left[\frac{2n+1}{6}-\frac{\left(n+1\right)}{4}\right]=\frac{\left(n+1\right)\left(n-1\right)}{12}=\frac{n^2-1}{12}$

And it is given that Variance$=14$

$\Rightarrow \frac{n^2-1}{12}=14$

$\Rightarrow n^2-1=168$

$\Rightarrow n=13$